Valuation Ring of P-adic Norm on Rationals

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Theorem

Let $\norm {\,\cdot\,}_p$ be the $p$-adic norm on the rationals $\Q$ for some prime $p$.


The induced valuation ring on $\struct {\Q,\norm {\,\cdot\,}_p}$ is the set:

$\OO = \Z_{\paren p} = \set {\dfrac a b \in \Q : p \nmid b}$


Corollary

The set of integers $\Z$ is a subring of $\OO$.

Proof

Let $\nu_p: \Q \to \Z \cup \set {+\infty}$ be the $p$-adic valuation on $\Q$.

Then:

\(\ds \OO\) \(=\) \(\ds \set {\dfrac a b \in \Q : \norm {\dfrac a b}_p \le 1}\) Definition of Valuation Ring Induced by Non-Archimedean Norm
\(\ds \) \(\) \(\ds \)
\(\ds \) \(=\) \(\ds \set{\dfrac a b \in \Q : \map {\nu_p} {\dfrac a b} \ge 0}\) Definition of P-adic Norm
\(\ds \) \(\) \(\ds \)
\(\ds \) \(=\) \(\ds \set {\dfrac a b \in \Q : \map {\nu_p} a - \map {\nu_p} b \ge 0}\) Definition of $p$-adic Valuation on Rationals
\(\ds \) \(\) \(\ds \)
\(\ds \) \(=\) \(\ds \set {\dfrac a b \in \Q : \map {\nu_p} a \ge \map {\nu_p} b}\)


Let $\dfrac a b \in \Q$ be in canonical form.

Then $a \perp b$


Suppose $p \divides b$.

Then $p \nmid a$.

Hence:

$\map {\nu_p} b \gt 0 = \map {\nu_p} a$


Suppose $p \nmid b$.

Then:

$\map {\nu_p} a \ge 0 = \map {\nu_p} b$


So:

$\map {\nu_p} a \ge \map {\nu_p} b$ if and only if $p \nmid b$

Hence:

$\OO = \set {\dfrac a b \in \Q : p \nmid b }$

$\blacksquare$


Sources

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